A soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest. The foot of the player is in contact with the ball for$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{s}$, and the force of the kick is given by

role="math" localid="1661489189931" $\mathit{F}\left(t\right)\mathbf{=}\left[\left(6.0\times {10}^6\right)t-\left(2.0\times {10}^9\right)t2\right]\mathbf{N}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$

Where t is in seconds. Find the magnitudes of the (a) impulse on the ball due to the kick, (b) The average force on the ball from the player’s foot during the period of contact, (c) The maximum force on the ball from the player’s foot during the period of contact, and (d) The ball’s velocity immediately after it loses contact with the player’s foot.

i) The mass of the ball m is 0.45 kg.

ii) The initial speed of the ball is 0 m/s.

iii) The period of contact with the ball t is 3.0 x 10-3 s .

iv) The force acting on the ball $F\mathrm{is}\left(6.0\times {10}^6\right)\mathrm{t}-\left(2.0\times {10}^9\right){\mathrm{t}}^2$ .

v) The time for which the force acts is $0\le t\le 3.0\times {10}^{-3}\mathrm{s}$.

The player kicks the ball. It imparts an impulse on the ball during the contact with the ball. We can use the impulse – linear momentum theorem to determine the impulse, average force and maximum force on the ball and given as.

$$\begin{gathered} \mathbf{J}\mathbf{=}\mathbf{∆}\mathbf{p} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{m}\mathbf{}\mathbf{∆}\mathbf{}\mathbf{v} \\ \mathbf{J}\mathbf{=}{\mathbf{\int }}_{{\mathbf{t}}_{\mathbf{1}}}^{{\mathbf{t}}_{\mathbf{2}}}\mathbf{Fdt} \end{gathered}$$

(a)

The impulse - linear momentum theorem can be stated as:

$\mathrm{J}={\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Fdt}$

Hence, the impulse can be calculated as

$$\begin{gathered} \mathrm{J}={\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}\mathrm{Fdt} \\ ={\int }_0^{3\times {10}^{-3}}\left(\left(6.0\times {10}^6\right)t-\left(2.0\times {10}^9\right)t^2\right)dt \\ ={\left[\left(6.0\times {10}^6\right)\frac{t^2}{2}-\left(2.0\times {10}^9\right)\frac{t^3}{3}\right]}_0^{3\times {10}^{-3}} \\ =\left[\left(3.0\times {10}^6\times 9\times {10}^{-6}\right)-\left(0.66\times {10}^9\times 27\times {10}^{-9}\right)\right]-\left(0\right) \\ =27-18 \\ =9.0\mathrm{N}.\mathrm{s} \end{gathered}$$

(b)

The average force can be calculated as

$$\begin{gathered} J=F_{\mathrm{avg}}∆t \\ =\frac{J}{∆t} \\ =\frac{9.0}{3\times {10}^{-3}} \\ =3\times {10}^3\mathrm{N} \\ =3.0\mathrm{kN} \end{gathered}$$

(c)

The force is maximum when

$$\begin{gathered} \frac{dF}{dt}=0 \\ \frac{dF}{dt}=\left(6.0\times {10}^6\right)-\left(2\times 2\times {10}^9\right)t \\ \frac{dF}{dt}=0 \\ =\left(6.0\times {10}^6\right)-\left(2\times 2\times {10}^9\right)t=0 \\ \left(2\times 2\times {10}^9\right)t=6.0\times {10}^6 \\ t=\frac{6.0\times {10}^6}{4.0\times {10}^9}=1.5\times {10}^{-3}\mathrm{s} \end{gathered}$$

Thus, force is zero at $1.5\times {10}^{-3}\mathrm{s}$

$$\begin{gathered} F_{max}=F\left(t=1.5\times {10}^{-3}\right)=\left(6.0\times {10}^6\right)\left(1.5\times {10}^{-3}\right)-\left(2\times {10}^9\right){\left(1.5\times {10}^{-3}\right)}^2 \\ {F}_{max}=9.0\times {10}^3-4.5\times {10}^3 \\ =4.5\times {10}^3\mathrm{N} \\ \mathrm{Thus},F_{max}=4.5\mathrm{kN} \end{gathered}$$

(d)

The ball’s velocity immediately after it loses contact with the foot of the player can be determined using the definition of impulse as follows

$$\begin{gathered} J=∆\mathrm{p} \\ =m∆v \\ ∆v=\frac{J}{m} \\ =\frac{9.0}{0.45} \\ =20\mathrm{m}/\mathrm{s} \end{gathered}$$

But the initial velocity is zero. Hence change in velocity is equal to final velocity of the ball

Thus, the velocity of the ball after it loses contact with the foot is 20 m/s.